example exercise 9.1 atomic mass and avogadro’s …example exercise 9.1 atomic mass and...

© 2011 Pearson Education, Inc.Introductory Chemistry: Concepts and Critical Thinking, 6th EditionCharles H. Corwin

Example Exercise 9.1 Atomic Mass and Avogadro’s Number

The atomic mass of each element is listed below the symbol of the element in the periodic table: Cu = 63.55 amu, Hg = 200.59 amu, S = 32.07 amu, and He = 4.00 amu. The mass of Avogadro’s number of atoms is the atomic mass expressed in grams. Therefore, 6.02 × 1023 atoms of(a) Cu = 63.55 g (c) S = 32.07 g(b) Hg = 200.59 g (d) He = 4.00 g

Solution

Refer to the atomic masses in the periodic table inside the front cover of this textbook. State the mass of Avogadro’s number of atoms for each of the following elements:(a) Copper (c) sulfur(b) Mercury (d) helium

Refer to the periodic table and state the mass for each of the following:(a) 1 atom of Au (b) 6.02 × 1023 atoms of Au

Answers: (a) 196.97 amu; (b) 196.97 g

Practice Exercise

What is the mass of an average platinum atom? What is the mass of Avogadro’s number of Pt atoms?

Answer: See Appendix G.

Concept Exercise


Example Exercise 9.2 Mole Calculations ICalculate the number of sodium atoms in 0.120 mol Na.

Strategy Plan

Step 1: What unit is asked for in the answer?

Step 2: What given value is related to the answer?

Step 3: What unit factor(s) should we apply?

Since 1 mol Na = 6.02 × 1023 atoms Na, the two unit factors are 1 mol Na/6.02 × 1023 atoms Na, and its reciprocal 6.02 × 1023 atoms Na/1 mol Na.

Unit Analysis Map


Example Exercise 9.2 Mole Calculations I

Calculate the number of formula units in 0.0763 mol of sodium chloride, NaCl.

Answers: 4.59 × 1022 formula units NaCl

Practice Exercise

What is the number of molecules in 1.00 mol of any gas?


Concept Exercise

We apply the unit factor 6.02 × 1023 atoms Na/1 mol Na to cancel moles , which appears in the denominator.

The answer is rounded to three digits because the given value and unit factor each have three significant digits.

SolutionContinued


Example Exercise 9.3 Mole Calculations ICalculate the number of moles of potassium in 1.25 × 1021 atoms K.

Strategy PlanStep 1: What unit is asked for in the answer?



Since mol K = 6.02 × 1023 atoms K, the two unit factors are 1 mol K/6.02 × 1023 atoms K, and its reciprocal 6.02 × 1023 atoms K/1 mol K.

Unit Analysis Map


Example Exercise 9.3 Mole Calculations I

We apply the unit factor 1 mol K/6.02 × 1023 atoms K to cancel atoms , which appears in the denominator.

The answer is rounded to three digits because the given value and unit factor each have three significant digits.

Solution

Calculate the number of moles of potassium iodide in 5.34 × 1025 formula units of KI.

Answer: 88.7 mol KI

Practice Exercise

What is the number of molecules in 1.00 mol of iodine crystals?


Concept Exercise

Continued


Example Exercise 9.4 Molar Mass Calculations

We begin by finding the atomic mass of each element in the periodic table. The molar mass equals the sum of the atomic masses expressed in g/mol.(a) The atomic mass of Ag is 107.87 amu, and the molar mass of silver equals 107.87 g/mol.(b) The sum of the atomic masses for NH3 is 14.01 amu + 3(1.01)amu = 17.04 amu. The molar mass of ammonia equals 17.04 g/mol.(c) The sum of the atomic masses for Mg(NO3)2 is 24.31 amu + 2(14.01 + 16.00 + 16.00 + 16.00)amu = 148.33 amu.The molar mass of magnesium nitrate equals 148.33 g/mol.

Solution

Calculate the molar mass for each of the following substances:(a) silver metal, Ag (b) ammonia gas, NH3 (c) magnesium nitrate, Mg(NO3)2

Calculate the molar mass for each of the following substances:(a) manganese metal, Mn (b) sulfur hexafluoride, SF6 (c) strontium acetate, Sr(C2H3O2)2

Answers: (a) 54.94 g/mol; (b) 146.07 g/mol; (c) 205.72 g/mol

Practice Exercise

The molecular mass of water is 18.02 amu. What is the mass of Avogadro’s number of water molecules?


Concept Exercise


Example Exercise 9.5 Mole Calculations IIWhat is the mass in grams of 2.01 × 1022 atoms of sulfur?




By definition, 1 mol S = 6.02 × 1023 atoms S, and the molar mass from the periodic table is 32.07 g S = 1 mol S; the two pairs of unit factors are shown in Step 3.

Strategy Plan

Unit Analysis Map


Example Exercise 9.5 Mole Calculations II

We apply the unit factor 1 mol S/6.02 × 1023 atoms S to cancel atoms S , and 32.07 g S/1 mol S to cancel mol S :

Solution

What is the mass of 7.75 × 1022 formula units of lead(II) sulfide, PbS?

Answer: 30.8 g PbS

Practice Exercise

Sulfur occurs naturally as S8 molecules. What is the mass of Avogadro’s number of S8 molecules?


Concept Exercise

Continued


Example Exercise 9.6 Mole Calculations IIHow many O2 molecules are present in 0.470 g of oxygen gas?




The molar mass from the periodic table is 32.00 g O2 = 1 mol O2, and by definition, 1 mol O2 = 6.02 × 1023 molecules O2. Thus, the two pairs of unit factors are shown in Step 3.

Unit Analysis Map

Strategy Plan


Example Exercise 9.6 Mole Calculations II

We apply the unit factor 1 mol O2/32.00 g O2 to cancel g O2 , and 6.02 × 1023 molecules O2/1 mol O2 to cancel moles O2 . Thus,

Solution

How many formula units of lithium fluoride are found in 0.175 g LiF?

Answer: 4.06 × 1021 formula units LiF

Practice Exercise

Ozone occurs naturally as O3 molecules. What is the mass of Avogadro’s number of ozone molecules?


Concept Exercise

Continued


Example Exercise 9.7 Mass of a MoleculeCalculate the mass in grams for a single molecule of carbon dioxide, CO2 (given that MM = 44.01 g/mol).




By definition, 1 mol CO2 = 6.02 × 1023 molecules. Thus, the unit factors are shown in Step 3.

Strategy Plan

Unit Analysis Map


Example Exercise 9.7 Mass of a Molecule

We apply the unit factor 1 mol CO2/6.02 × 1023 molecules CO2 to cancel mol CO2 .

Solution

Calculate the mass in grams for a single molecule of carbon monoxide, CO.

Answer: 4.65 × 10-23 g/molecule

Practice Exercise

What is the mass of a carbon monoxide molecule expressed in amu?


Concept Exercise

Continued


Example Exercise 9.8 Density of a Gas at STPCalculate the density of ammonia gas, NH3, at STP. (MM = 17.04 g/mol)




From the molar volume of a gas at STP, 1 mol = 22.4 L; thus, the unit factors are 1 mol/22.4 L, and 22.4 L/1 mol.

Strategy Plan

Unit Analysis Map


Example Exercise 9.8 Density of a Gas at STP

We apply the unit factor 1 mol /22.4 L to cancel moles , which appears in the denominator.

If we filled a balloon with ammonia gas (0.761 g/L), the balloon would float in air (1.29 g/L). Similarly, a less dense liquid floats on a more dense liquid; for example, gasoline (0.7 g/mL) floats on water (1.0 g/mL).

Solution

Calculate the density of hydrogen sulfide gas, H2S, at STP.

Answer: 1.52 g/L

Practice Exercise

Why does the balloon filled with helium gas rise, while the balloon filled with nitrogen gas does not?


Concept Exercise

Continued


Example Exercise 9.9 Molar Mass of a GasA fire extinguisher releases 1.96 g of an unknown gas that occupies 1.00 L at STP. What is the molar mass (g/mol) of the unknown gas?




From the molar volume of a gas at STP, 1 mol = 22.4 L; thus, the unit factors are 1 mol/22.4 L, and 22.4 L/1 mol.

Strategy Plan

Unit Analysis Map


Example Exercise 9.9 Molar Mass of a Gas

We apply the unit factor 1 mol/22.4 L to cancel liters , which appears in the denominator.

Because the unknown gas is from a fire extinguisher, we suspect that it is carbon dioxide. By summing the molar mass for CO2(44.01 g/mol), we help confirm that the unknown gas is carbon dioxide.

Solution

Boron trifluoride gas is used in the manufacture of computer chips. Given that 1.505 g of the gas occupies 497 mL at STP, what is the molar mass of boron fluoride gas?

Answer: 67.8 g/mol

Practice Exercise

What is the volume of one mole of any gas at STP?


Concept Exercise

Continued


What is the mass of 3.36 L of ozone gas, O3, at STP?




From the molar volume of a gas at STP, 1 mol O3 = 22.4 L O3. The molar mass from the periodic table is 48.00 g O3 = 1 mol O3. Thus, the two pairs of unit factors are shown in Step 3.

Example Exercise 9.10 Mole Calculations III

Strategy Plan

Unit Analysis Map


What volume is occupied by 0.125 g of methane gas, CH4, at STP?

Answer: 0.174 L CH4 (174 mL)

Practice Exercise

We apply the unit factor 1 mol O3/22.4 L O3 to cancel L O3 , and 48.00 g O3/1 mol O3 to cancel moles O3 ,

How many molecules of ozone are in one mole of gas?


Concept Exercise


Solution

Continued


Example Exercise 9.11 Mole Calculations IIIHow many molecules of hydrogen gas, H2, occupy 0.500 L at STP?




From the molar volume of a gas at STP, 1 mol H2 = 22.4 L H2 and by definition, 1 mol H2 = 6.02 × 1023 molecules H2. Thus, the two pairs of unit factors are shown in Step 3.

Strategy Plan

Unit Analysis Map



We apply the unit factor 1 mol H2/22.4 L H2 to cancel L H2 , and 6.02 × 1023 molecules H2/1 mol H2to cancel moles 77H2 . Thus,

Solution

What is the volume occupied by 3.33 × 1021 atoms of helium gas, He, at STP?

Answer: 0.124 L He (124 mL)

Practice Exercise

What is the mass of one mole of hydrogen gas?


Concept Exercise

Continued


Example Exercise 9.12 Percent Composition of a Substance

Let’s calculate the percent composition assuming there is 1 mol of TNT. For compounds with parentheses, it is necessary to count the number of atoms of each element carefully. That is, 1 mol of C7H5(NO2)3 contains 7 mol of C atoms, 5 mol of H atoms, 3 mol of NO2 (that is, 3 mol N atoms and 6 mol of O atoms).We begin the calculation by finding the molar mass of C7H5(NO2)3 as follows:

Now, let’s compare the mass of each element to the total molar mass of the compound, that is, 227.15 g.

The percent composition of TNT reveals that the explosive is mostly oxygen by mass. When we sum the individual percentages (37.01 + 2.22 + 18.50 + 42.26 = 99.99%), we verify our calculation and, find that the total is approximately 100%.

Solution

TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.


Example Exercise 9.12 Percent Composition of a Substance

EDTA (ethylenediaminetetraacetic acid) is used as a food preservative and in the treatment of metallic lead poisoning. Calculate the percent composition of EDTA, C10H16N2O8.

Answers: 41.09% C, 5.53% H, 9.587% N, and 43.79% O

Practice Exercise

If an analysis of sugar, CxHyOz, gave 40.0% C and 6.7% H, what is the percent oxygen?

Answers: See Appendix G.

Concept Exercise

Continued


Example Exercise 9.13 Empirical Formula from Mass Composition

The empirical formula is the simplest whole-number ratio of scandium and oxygen in the compound scandium oxide. This ratio is experimentally determined from the moles of each reactant. The moles of scandium are calculated as follows:

The moles of oxygen are calculated after first subtracting the mass of Sc from the product, Sc?O?:

The moles of oxygen are calculated from the mass of oxygen that reacted:

The mole ratio in scandium oxide is Sc0.0111O0.0167. To simplify the ratio and obtain small whole numbers, we divide by the smaller number:

Since the calculated ratio is not close to whole numbers, we cannot round off the experimental ratio Sc1.00O1.50. However, we can double the ratio to obtain whole numbers, that is, Sc2.00O3.00. Thus, the empirical formula is Sc2O3.

Solution

In a laboratory experiment, 0.500 g of scandium was heated and allowed to react with oxygen from the air. The resulting product oxide had a mass of 0.767 g. Now, let’s find the empirical formula for scandium oxide, Sc?O?.


Example Exercise 9.13 Empirical Formula from Mass Composition

Iron can react with chlorine gas to give two different compounds, FeCl2 and FeCl3. If 0.558 g of metallic iron reacts with chlorine gas to yield 1.621 g of iron chloride, which iron compound is produced in the experiment?

Answer: FeCl3

Practice Exercise

If 0.500 mol of yellow powder sulfur reacts with 0.500 mol of oxygen gas, what is the empirical formula of the sulfur oxide?


Concept Exercise

Continued


Example Exercise 9.14 Empirical Formula from Percent Composition

If we assume a 100 g sample, then the percentage of each element equals its mass in 100 g of glycine, that is, 32.0 g C, 6.7 g H, 18.7 g N, and 42.6 g O. We can determine the empirical formula as follows:

The mole ratio of the elements in the amino acid is C2.66H6.6N1.33O2.66. We can find a small whole-number ratio by dividing by the smallest number:

Simplifying, we find that the empirical formula for the amino acid glycine is C2H5NO2.

Solution

Glycine is an amino acid found in protein. An analysis of glycine gave the following data: 32.0% carbon, 6.7% hydrogen, 18.7% nitrogen, and 42.6% oxygen. Calculate the empirical formula of the amino acid.


Example Exercise 9.14 Empirical Formula from Percent Composition

Calculate the empirical formula for caffeine given the following percent composition: 49.5% C, 5.15% H, 28.9% N, and 16.5% O.

Answer: C4H5N2O

Practice Exercise

If 0.500 mol of yellow powder sulfur reacts with 0.750 mol of oxygen gas, what is the empirical formula of the sulfur oxide?


Concept Exercise

Continued


Example Exercise 9.15 Molecular Formula from Empirical Formula

We can indicate the molecular formula of fructose as (CH2O)n. The molar mass of the empirical formula CH2O is 12 g C + 2(1 g H) + 16 g O = 30 g/mol. Thus, the number of multiples of the empirical formula is

Thus, the molecular formula of fructose is (CH2O)6 or C6H12O6.

Solution

The empirical formula for fructose, or fruit sugar, is CH2O. If the molar mass of fructose is 180 g/mol, find the actual molecular formula for the sugar.

Ethylene dibromide was used as a grain pesticide until it was banned. Calculate (a) the empirical formula and (b) the molecular formula for ethylene dibromide given its approximate molar mass of 190 g/mol and its percent composition: 12.7% C, 2.1% H, and 85.1% Br.

Answers: (a) CH2Br; (b) C2H4Br2

Practice Exercise

If the molecular formula of hydrogen peroxide is H2O2, what is the empirical formula?


Concept Exercise

example exercise 9.1 atomic mass and avogadro’s …example exercise 9.1 atomic mass and...

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